I was told by one of my friends that if I wanted to get rid of the "city on fire" effect I had to use a very long exposure and high f-stop. I'm not sure how that exactly works but thats what they were telling me, and it seems to work for his camera to do it that way.
this image was done at fstop 8, and at a 15 second exposure.
And this other picture was done at an fstop of 3.5 or whatever the lowest setting was, at a 1 second exposure.
My friend pointed out that in the right side there, the city scape, in the area where Harbor island is... looks like it's on fire from all the light glare. And I was attempting to get rid of that.
Actually I think at the fstop of 8 it did a pretty good job on that first image there. I didn't get whole lot of light glare with that one.
The "City on Fire" effect is simply overexposure. A challenge of night photography is often very high contrast. That is the ratio between the darkest region and the brightest region is very high, at least when you are photographing scenes where artificial lighting is present.
Film has only so much exposure lattitude and the same is true of digital. Only so many values can be encoded in however many bits are allocated to a pixel. With very high contrast situations you can have a situation where the contrast of the scene exceeds the lattitude of the medium. In these cases, you are either going to have some portion of the scene overexposed and lose highlight detail or you are going to have some portion underexposed and lose shadow detail, or possibly both.
With film you can choose low contrast films with wide exposure latitudes for high contrast scenes. With digital, unless your camera has a setting for contrast (and I thought the Minolta did), you're stuck with what you've got and the best you can do is find a good compromise. In your first photo, you are losing shadow detail, you can't see the clouds or water except where bright lights reflect off of it. In the second, you are losing highlight details.
When contrast is that high it is impossible to reproduce the scene faithfully, you're going to lose some details on the highlights or shadows or both. Instead consider the asthetics. What I usually do is bracket widely and then later pick the image I think looks the best.
With respect to f/stop and exposure time, think of filling a bucket. The digital imager or film is the bucket. You can fill the bucket with a hose just dribbling water over a long period of time, or you can fill it in a short time with the hose on full blast. What is important is the total volume of water. Same for light, for a given exposure you can use a high f/stop (only let a dribble of light through) and a long exposure time, or a low f/stop (let a lot of light through) for a short time. It is the product of the rate at which light is allowed through and the amount of time that it is allowed through that determines the total exposure.
F3.5 will allow 5.34 times as much light through as F8, so a one-second exposure at F3.5 is equivalent to 5.34 second exposure at F8, so something is wrong with your numbers above, the 15 second F8 exposure should be about 300% more exposed than the the 1 second exposure at F3.5.
However, there is an additional factor complicating this and that is the fact that focal length is clearly not the same in the two images above. Assuming they had been the same, this is how you can understand and calculate the effect of f/stop on exposure.
For a given focal length, the amount of light that will reach the film or digital imager is a function of the area of the aperture. The f/stop is actually the focal length divided by the aperture diameter, if the aperture was at the front of the lens, usually it is not so that mucks up the calculation if you looked at the real diameter of the aperature, but the value printed on the lens or that you see when you set it on the digital camera is as if it were up front.
So lets say you've got a 55mm lens. An f/stop of 3.5 would mean the ratio of the focal length (55mm) to aperature diameter is 3.5, therefor the aperature diameter is 15.71mm. At F8, the ratio of focal length to diameter is 8, so the aperature diameter would be 6.8mm. The aperature approximates a circle. The area of a circle is pi r^2, the radius is half of the diameter. So at F3.5, the aperature area is 3.14 * 7.86^2 or 193.74 square mm. At F8, 3.14 * 3.40^2 or 36.30 quare mm. 193.74 / 36.30 = 5.34, actually calculated at higher precision I get 5.22, but close enough. Anyway, that's how you can calculate the relative exposure of two different f/stop values. The total exposure being the product of aperature area and exposure time, 1 second at F3.5 is equivalent to 5.34 seconds at F8.
With digital this will be accurate. Film suffers to varying degrees to what is termed repricocity failure which means it acts like a lower speed film with long exposures so the exposure time * aperature area doesn't strictly hold with long exposures.
Back to digital, now that you know you can get the same exposure with a high f/stop and long exposure or a low f/stop and a short exposure, there are other things to consider. A high f/stop gives a wide depth of field, that is you can have things far away and relatively close both relatively in focus. With a low f/stop, you get a narrow depth of field, getting closer or farther away from the distance where things are completely in focus, things will be out of focus.
The other thing to consider is motion blur, a longer exposure, necessitated by a higher f/stop, results in more motion blur. Say you are taking a picture that includes the freeway. Cars are moving at around 60 mph, or about 88 feet/second. So if you take a one second exposure you're going to get light streaks 88 feet long, if you take a 15 second exposure, the streaks will be 1320 feet, or a quarter mile long. Water and clouds move, do you want to freeze the waves or create a blur? Clouds move, do you want to see cloud detail or a blur?
One more variable you have is film speed which is tweakable on most digital cameras. Just like real film, higher speeds are "grainier", that is there is more noise in the image. Unlike film, higher speeds on digital does not degrade color, contrast, or resolution. Higher speed film uses larger grains which results in lower resolution and lower loss chromatic filters resulting in lower color, and generally contrast is lower as well.
One other thing to consider with a digital camera is that you can do some post processing. What I have found with the Minolta is that when the highlights appear saturated, they truely are. But when the shadow detail appears lost, it is in fact still there and you can stretch out the low end of the dynamic range with the Minolta image processing software. So what I would do in this situation is go with the lower exposure and then stretch out the low end of the brightness range to recover some shadow detail.
Anyway there ya go, there is some science involved. Don't let it get in the way of asthetics, but being aware of it can help you achieve the asthetics you desire.